General Values of Inverse Trigonometric Functions

We will learn how to find the general values of inverse trigonometric functions in different types of problems.

1. Find the general values of sin\(^{-1}\) (- √3/2)

Solution:  

Let, sin\(^{-1}\) (- √3/2) = θ

Therefore, sin θ = - √3/2

⇒ sin θ = - sin (π/3)

⇒ sin θ = (- π/3)

Therefore, the general value of sin\(^{-1}\) (- √3/2) = θ = nπ - (- 1)\(^{n}\) π/3, where, n = 0 or any integer.

2. Find the general values of cot\(^{-1}\) (- 1)

Solution:

Let, cot\(^{-1}\) (- 1) = θ                    

Therefore, cot θ = - 1

⇒ cot θ = cot (- π/4)

Therefore, the general value of cot\(^{-1}\) (- 1) = θ = nπ - π/4, where, n = 0 or any integer.

 

3. Find the general values of cos\(^{-1}\) (1/2)      

Solution:   

Let, cos\(^{-1}\) 1/2 = θ           

Therefore, cos θ = 1/2

⇒ cos θ = cos (π/3)

Therefore, the general value of cos\(^{-1}\) (1/2) = θ = 2nπ ± π/3, where, n = 0 or any integer.

 

4. Find the general values of sec\(^{-1}\) (- 2)  

Solution:

Let, sec\(^{-1}\) (- 2) = θ

Therefore, sec θ = - 2

⇒ sec θ = - sec (π/3)

⇒ sec θ = sec (π - π/3)

⇒ sec θ = sec (2π/3)

Therefore, the general value of sec\(^{-1}\) (- 2) = θ = 2nπ ± 2π/3, where, n = 0 or any integer.

 

5. Find the general values of csc\(^{-1}\) (√2)

Solution:

Let, csc\(^{-1}\) (√2) = θ           

Therefore, csc θ = √2 .

⇒csc θ = csc (π/4)

Therefore, the general value of csc\(^{-1}\) (√2 ) = θ = nπ + (- 1)\(^{n}\) π/4, where, n = 0 or any integer.

 

6. Find the general values of tan\(^{-1}\) (√3)

Solution:

 Let, tan\(^{-1}\) (√3) = θ                   

Therefore, tan θ = √3

⇒ tan θ = tan (π/3)

Therefore, the general value of tan\(^{-1}\) (√3) = θ = nπ + π/3 where, n = 0 or any integer.

 Inverse Trigonometric Functions






11 and 12 Grade Math

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